Problem: $\vec v = (3,-4)$ $5\vec v= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $5 \vec{v}$ : $\begin{aligned} {5}\vec v = {5} \cdot (3,-4) &= \left({5} \cdot 3, {5} \cdot (-4)\right) \\\\ &= (15,-20) \end{aligned}$ The answer is $ (15,-20) $.